A162319 Array read by antidiagonals: a(n,m) = the number of digits of m is when written in base n. The top row is the number of digits for each m in base 1.
1, 1, 2, 1, 2, 3, 1, 1, 2, 4, 1, 1, 2, 3, 5, 1, 1, 1, 2, 3, 6, 1, 1, 1, 2, 2, 3, 7, 1, 1, 1, 1, 2, 2, 3, 8, 1, 1, 1, 1, 2, 2, 2, 4, 9, 1, 1, 1, 1, 1, 2, 2, 2, 4, 10, 1, 1, 1, 1, 1, 2, 2, 2, 3, 4, 11, 1, 1, 1, 1, 1, 1, 2, 2, 2, 3, 4, 12, 1, 1, 1, 1, 1, 1, 2, 2
Offset: 1
Examples
From _Michael De Vlieger_, Jan 02 2015: (Start)
Array read by antidiagonals begins:
1;
1, 2;
1, 2, 3;
1, 1, 2, 4;
1, 1, 2, 3, 5;
1, 1, 1, 2, 3, 6;
1, 1, 1, 2, 2, 3, 7;
1, 1, 1, 1, 2, 2, 3, 8;
1, 1, 1, 1, 2, 2, 2, 4, 9;
1, 1, 1, 1, 1, 2, 2, 2, 4, 10;
...
Array adjusted such that the rows represent base n and the columns m:
m
1 2 3 4 5 6 7 8 9 10
------------------------------
base 1: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10;
base 2: 1, 2, 2, 3, 3, 3, 3, 4, 4, (4);
base 3: 1, 1, 2, 2, 2, 2, 2, 2, (3, 3);
base 4: 1, 1, 1, 2, 2, 2, 2, (2, 2, 2);
base 5: 1, 1, 1, 1, 2, 2, (2, 2, 2, 2);
base 6: 1, 1, 1, 1, 1, (2, 2, 2, 2, 2);
base 7: 1, 1, 1, 1, (1, 1, 2, 2, 2, 2);
base 8: 1, 1, 1, (1, 1, 1, 1, 2, 2, 2);
base 9: 1, 1, (1, 1, 1, 1, 1, 1, 2, 2);
base 10: 1, (1, 1, 1, 1, 1, 1, 1, 1, 1);
...
For n = 12, a(12) is found in the second position in row 5 in the array read by antidiagonals. This equates to m = 2, base n = 4. The number m = 2 in base n = 4 requires 1 digit, thus a(12) = 1.
For n = 14, a(14) is found in the fourth position in row 5 in the array read by antidiagonals. This equates to m = 4, base n = 2. The number m = 4 in base n = 2 requires 3 digits, thus a(14) = 3. (End)
Links
- Michael De Vlieger, Table of n, a(n) for n = 1..10440 (covers bases 1..144)
Crossrefs
Cf. A162320.
Programs
-
Mathematica
Table[Function[k, If[k == 1, m, IntegerLength[m, k]]][k - m + 1], {k, 13}, {m, k}] // Flatten (* Michael De Vlieger, Aug 31 2017 *)
Comments